ALMOST PRINCIPALLY SMALL INJECTIVE RINGS 1. Introduction Let

J. Korean Math. Soc. 48 (2011), No. 6, pp. 1189–1201 http://dx.doi.org/10.4134/JKMS.2011.48.6.1189

ALMOST PRINCIPALLY SMALL INJECTIVE RINGS Yueming Xiang Abstract. Let R be a ring and M a right R-module, S = EndR (M ). The module M is called almost principally small injective (or AP Sinjective for short) if, for any a ∈ J(R), there exists an S-submodule Xa of M such that lM rR (a) = M a ⊕ Xa as left S-modules. If RR is an AP S-injective module, then we call R a right AP S-injective ring. We develop, in this paper, AP S-injective rings as a generalization of P Sinjective rings and AP -injective rings. Many examples of AP S-injective rings are listed. We also extend some results on P S-injective rings and AP -injective rings to AP S-injective rings.

1. Introduction Let R be a ring. A right ideal I of R is called small if, for every proper right ideal K of R, K +I ̸= R. Recall that a ring R is right principally small injective (or P S-injective) (resp. P -injective, small injective, mininjective) if every Rhomomorphism f : I → R, for every principally small (resp. principally, small, minimal) right ideal I, can be extended to R. The detailed discussion of P injective, small injective and mininjective rings can be found in [2, 3, 4, 8, 9, 10, 12]. The concept of P S-injective rings was first introduced in [14] as a generalization of P -injective rings and small injective rings. It was shown that every right P S-injective ring is also right mininjective. In [11], Page and Zhou introduced AP -injectivity and AGP -injectivity of modules and rings. Given a right R-module M , S = EndR (M ). The module M is called AP -injective if, for any a ∈ R, there exists an S-submodule Xa of M such that lM rR (a) = M a⊕Xa as left S-modules. The module M is called AGP -injective if, for any 0 ̸= a ∈ R, there exists a positive integer n = n(a) and an S-submodule Xa of M such that an ̸= 0 and lM rR (an ) = M an ⊕ Xa as left S-modules. A ring R is called right AP -injective (resp. AGP -injective) if RR is an AP -injective (resp. AGP injective) module. Many of the results on right P -injective rings were obtained for the two classes of right AP -injective rings and right AGP -injective rings. In [17], Zhou continued the study of left AP -injective rings and left AGP -injective rings with various chain conditions. Received June 19, 2010. 2010 Mathematics Subject Classification. 16N20, 16P40, 16D50. Key words and phrases. AP S-injective modules (rings), trivial extensions. c ⃝2011 The Korean Mathematical Society

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In the present paper, we say that a right R-module M is AP S-injective if, for any a ∈ J(R), there exists an S-submodule Xa of M such that lM rR (a) = M a ⊕ Xa as left S-modules. A ring R is called right AP S-injective if RR is an AP S-injective module. Similarly, we can define a left AP S-injective ring. Some examples are listed to show that AP S-injective rings are the proper generalization of P S-injective rings and AP -injective rings. It is also shown that there are many similarities between AP -injective rings and AP S-injective rings. In light of this fact, some results on P S-injective rings and AP -injective rings are as the corollaries of our results, respectively. Throughout R is an associative ring with identity and all modules are unitary. J = J(R), soc(RR ) and Z(RR ) denote the Jacobson radical, right socle and right singular ideal of R, respectively. For a right R-module M , let S = EndR (M ), then we have an (S, R)-bimodule M . If X is a subset of R, the right (left) annihilator of X in R is denoted by rR (X) (lR (X)). We write a ∈ L − I to indicate that a ∈ L but a ∈ / I and N ≤e M to indicate that N is an essential submodule of M . The notation M n stands for the direct sum of n copies of the module M , written as column matrices. For the usual notations we refer the reader to [1], [6] and [10]. 2. Examples and basic properties Definition 2.1. Let M be a right R-module, S = EndR (M ). The module M is called almost principally small injective (or AP S-injective for short) if, for any a ∈ J(R), there exists an S-submodule Xa of M such that lM rR (a) = M a⊕Xa as left S-modules. If RR is an AP S-injective module, then we call R a right AP S-injective ring. Similarly, we can define the concept of left AP S-injective rings. For an R-module N and a submodule P of N , we will identify HomR (N, M ) with the set of homomorphisms in HomR (P, M ) that can be extended to N , and hence HomR (N, M ) can be seen as a left S-submodule of HomR (P, M ). Lemma 2.2. Let MR be a module, S = EndR (M ) and a ∈ J(R). (1) If lM rR (a) = M a ⊕ X for some X ⊆ M as left S-modules, then HomR (aR, M ) = HomR (R, M ) ⊕ Γ as left S-modules, where Γ = {f ∈ HomR (aR, M ) : f (a) ∈ X}. (2) If HomR (aR, M ) = HomR (R, M ) ⊕ Y as left S-modules, then lM rR (a) = M a ⊕ X as left S-modules, where X = {f (a) : f ∈ Y }. (3) M a is a direct summand of lM rR (a) as left S-modules if and only if HomR (R, M ) is a direct summand of HomR (aR, M ) as left S-modules. Proof. The proof is similar to that of [11, Lemma 1.2].

□

From Lemma 2.2, we have the following corollary. Corollary 2.3. Let MR be a module and a ∈ J(R). Then lM rR (a) = M a if and only if every R-homomorphism of aR into M extends to R.

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Remark 2.4. (1) Obviously, right P S-injective modules are right AP S-injective. But the converse is false in general. For example, let R = ( F0 F F ) with F a field and MR = ( 00 F ). Then M is right AP S-injective but not right P S-injective. F In fact, choose 0 ̸= x ∈ F . Then a = ( 00 x0 ) ∈ J(R) and lM rR (a) = M ̸= M a = 0. By the preceding corollary, M is not right P S-injective. Note that J(R) = ( 00 F0 ). Thus, lM rR (a) = M a ⊕ M for any a ∈ J(R). Therefore, M is right AP S-injective. (2) Right AP -injective modules are right AP S-injective. (3) Right AP S-injective rings are right almost mininjective [13] (A ring R is called right almost mininjective if, for any minimal right ideal kR of R, there exists an S-submodule Xk of R such that lR rR (k) = Rk ⊕ Xk as left S-modules). In fact, in view of [6, Lemma 10.22], every minimal right ideal of R is either nilpotent or a direct summand of R. Example 2.5. The three examples of [11, Examples 1.5] are commutative AP S-injective but not P S-injective. Example 2.6. Let R = Z be the ring of integers. Then R is AP S-injective but not AGP -injective. Example 2.7. Let K be a field and L be a proper subfield of K such that ρ : K → L is an isomorphism, e.g., let K = F (y1 , y2 , . . .) with F a field, ρ(yi ) = yi+1 and ρ(c) = c for all c ∈ F . Let K[x1 , x2 ; ρ] be the ring of twisted right polynomials over K where kxi = xi ρ(k) for all k ∈ K and for i = 1, 2. Set R = K[x1 , x2 ; ρ]/(x21 , x22 ). In view of [3, Example 1 and Proposition 1], R is a left AGP -injective but not AP S-injective. Theorem 2.8. Let R be a right AP S-injective ring. Then. (1) J(R) ⊆ Z(RR ). (2) soc(RR ) ⊆ rR (J). Proof. (1) Take any a ∈ J(R). If a ∈ / Z(RR ), then there exists a nonzero right ideal I of R such that rR (a) ∩ I = 0. So there exists b ∈ I such that ab ̸= 0. Note that ab ∈ J(R), by hypothesis, there exists 0 ̸= u ∈ abR such that lR rR (u) = Ru ⊕ Xu , where Xu ⊆ R R. Write u = abc for some c ∈ R. If t ∈ rR (abc), then abct = 0, implying ct ∈ rR (ab) = rR (b) since rR (a) ∩ I = 0. Hence, (bc)t = b(ct) = 0, and so t ∈ rR (bc). This shows that rR (bc) = rR (abc). Note that bc ∈ lR rR (bc) = lR rR (abc) = Ru ⊕ Xu . Write bc = dabc + x, where dabc ∈ Ru − Xu and x ∈ Xu − Ru. Then x = (1 − da)bc, and so bc = (1 − da)−1 x ∈ Xu since 1 − da is invertible, contradicting with dabc ∈ Ru − Xu . (2) Let kR be a simple right ideal of R. Suppose jk ̸= 0 for some j ∈ J(R), then rR (jk) = rR (k). Note that jk ∈ J(R) and R is right AP S-injective. Then there exists a left ideal Xjk of R such that lR rR (jk) = Rjk ⊕ Xjk . Since k ∈ lR rR (jk), write k = rjk + x, where rjk ∈ Rjk − Xjk and x ∈ Xjk − Rjk. Then x = (1 − rj)k, and hence k = (1 − jk)−1 x ∈ Xjk since 1 − jk is invertible, contradicting with rjk ∈ Rjk − Xjk . □

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The following example shows that a right mininjective ring need not be right AP S-injective. Example 2.9. Let R be the ring of all N-square upper triangular matrices over a field F that are constant on the diagonal and have only finitely many nonzero entries oﬀ the diagonal. By [16, Example 1.7], soc(RR ) = Z(RR ) = 0 and J(R) ̸= 0. So R is right mininjective. However, R is not right AP Sinjective by Theorem 2.8(1). A ring R is called semiregular if R/J(R) is von Neumann regular and idempotents lift modulo J(R), equivalently if, for any a ∈ R, there exists e2 = e ∈ Ra such that a(1 − e) ∈ J(R) (cf. [10, Lemma B.40]). Proposition 2.10. If R is semiregular, then R is right AP -injective if and only if R is right AP S-injective. Proof. It is enough to prove suﬃcient condition. Since R is semiregular, for any a ∈ R, Ra = Re ⊕ Rb, where e2 = e ∈ R and b ∈ J(R). By hypothesis, lR rR (b) = Rb ⊕ Xb for some left ideal Xb of R. Then Ra ⊕ Xb = Re ⊕ Rb ⊕ Xb = lR (1 − e) ⊕ lR rR (b) = lR ((1 − e)R ⊕ rR (b)) = lR (rR (Re) ⊕ rR (Rb)) = lR rR (Re ⊕ Rb) = lR rR (Ra) = lR rR (a). Therefore, R is right AP -injective. □ Remark 2.11. There ( exists ) a ring that is semiregular but not right AP Sinjective. Let R = Z02 ZZ22 , where Z2 is the ring of integers modulo 2. Then ( ) J(R) = 00 Z02 , and Z(RR ) = 0. By Theorem 2.8, R is not right AP S-injective. ( ) But R/J(R) ∼ = Z02 Z02 is von Neumann regular and any idempotent of R/J(R) can be lifted to R, so R is semiregular. By Proposition 2.10 and [11, Theorem 2.16], we have the following result. Corollary 2.12. If R is a semiperfect and right AP S-injective ring, then R = R1 × R2 , where R1 is semisimple and every simple right ideal of R2 is nilpotent. Clearly, a semiprimitive ring (J(R) = 0) is left and right AP S-injective. But the converse is not true as Example 2.5. Next, we shall consider when a right AP S-injective ring is semiprimitive. Following [7], A ring R is called a right J − P P ring if aR is projective for any a ∈ J(R). Proposition 2.13. Let R be a ring. Then the following are equivalent: (1) R is semiprimitive. (2) R is right J − P P and right AP S-injective. (3) R is a right AP S-injective ring whose every simple singular right Rmodule is P S-injective. Proof. (1)⇒(2) and (1)⇒(3) are trivial. (2)⇒(1). Suppose 0 ̸= a ∈ J(R). Since R is right J − P P , aR is projective. So the exact sequence 0 → rR (a) → R → aR → 0 splits. Then rR (a) = eR for some e2 = e ∈ R. It follows that lR rR (a) = lR (eR) = R(1 − e). Note that


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R is also right AP S-injective, so there exists a left ideal Xa of R such that lR rR (a) = Ra ⊕ Xa . Then Ra is a direct summand of R(1 − e), and hence a direct summand of R R, which implies ∩ a = 0, a contradiction. ∩ (3)⇒(1). We first show that J Z(RR ) = 0. Take any b ∈ J Z(RR ). If b ̸= 0, then rR (b) + RbR is an essential right ideal of R. If rR (b) + RbR ̸= R, there exists a maximal essential right ideal T of R such that rR (b) + RbR ⊆ T . By hypothesis, R/T is P S-injective. Note that rR (b) ⊆ T , then the Rhomomorphism f : bR → R/T by br 7→ r + T is well defined. So f = (c + T )· for some c ∈ R. Then f (b) = 1 + T = cb + T . Note that cb ∈ RbR ⊆ T , so 1 ∈ T , a contradiction. This proves that rR (b) + RbR = R, and hence rR (b) = R because RbR is a small ideal of R. This implies b = 0, which is required contradiction. Therefore, J(R) = J ∩ Z(RR ) = 0 by Theorem 2.8(1). □ Now we construct a right AP S-injective ring that is not left AP S-injective. Example 2.14. Let R = ( K 0

K A

) , where K = Z2 and

A = {(a1 , a2 , . . . , an , a, a, . . .) | a, a1 , a2 , . . . ∈ K, n ∈ N}. If k ∈ K and (a1 , a2 , . . . , an , a, a, . . .) ∈ A, let k · (a1 , a2 , . . . , an , a, a, . . .) = ka. Following [2, Example 1], R is right P -injective, and hence right AP Sinjective. But J(R) = ( 00 K 0 ) ̸= 0, so R is not semiprimitive. We claim that R is not left AP S-injective. By Proposition 2.13, it is enough to show that ( ) every K K simple singular left R-module is P S-injective. In fact, M = 0 Z(N) is the 2 unique maximal essential right ideal of R, where (N)

Z2

= {(a1 , a2 , . . . , an , 0, 0, . . .) | a1 , a2 , . . . ∈ K, n ∈ N}.

In view of [15, p. 5], R = R/M is left P -injective, and hence left P S-injective. Proposition 2.15. If R is a right AP S-injective ring and R/soc(RR ) satisfies the ACC on right annihilators, then J(R) is nilpotent. Proof. Write S = soc(RR ) and R = R/S. For any sequence a1 , a2 , a3 , . . . ∈ J(R), there is an ascending chain rR (a1 ) ⊆ rR (a2 a1 ) ⊆ rR (a3 a2 a1 ) ⊆ · · · , by hypothesis, there exists a positive integer m such that rR (am · · · a2 a1 ) = rR (am+k · · · am · · · a2 a1 ), k = 1, 2, . . . . Since an+1 an · · · a1 ∈ J(R) ⊆ Z(RR ) by Theorem 2.8(1), rR (an+1 an · · · a1 ) is the essential right ideal of R. Then S ⊆ rR (an+1 an · · · a1 ). Now we prove that (1)

rR (an · · · a2 a1 ) ⊆ rR (an+1 an · · · a1 )/S ⊆ rR (an+1 an · · · a1 ).

In fact, for any b + S ∈ rR (an · · · a2 a1 ), an · · · a1 b ∈ S. Then an+1 an · · · a1 b = 0 because S ⊆ rR (an+1 ). So b ∈ rR (an+1 an · · · a1 ), and hence b + S ∈ rR (an+1 an · · · a1 )/S. But the second inclusion is clear.

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Since rR (am · · · a2 a1 ) = rR (am+2 am+1 · · · a2 a1 ), by (1), rR (am+1 am · · · a1 )/S = rR (am+2 am+1 · · · a1 )/S. Then rR (am+1 am · · · a1 ) = rR (am+2 am+1 · · · a1 ), and so (am+1 am · · · a1 )R ∩ rR (am+2 ) = 0. Note that rR (am+2 ) is also an essential right ideal of R, then am+1 am · · · a1 = 0. So J(R) is a right T -nilpotent ideal and the ideal J(R) + S/S of R is also a right T -nilpotent. By [1, Proposition 29.1], J(R) + S/S is nilpotent. Then there exists a positive integer t such that (J(R))t ⊆ S, so (J(R))t+1 ⊆ J(R)S = 0, as desired. □ Proposition 2.16. If R is a right AP S-injective (resp. P S-injective, AP injective) ring, so is eRe for all e2 = e ∈ R such that ReR = R. Proof. Let S = eRe and let a ∈ J(S) = eJe. Then a = ae ∈ J(R), so there exists a left ideal Xa of R such that lR rR (a) = Ra ⊕ Xa . Since 1 − e ∈ rR (a), we see that t(1 − e) = 0 for any t ∈ Xa , which implies Xa = Xa e. Thus eRae∩eXa e = 0. Clearly, eRae ⊆ lS rS (a) and eXa e ⊆ lS rS (a) since Rae = Ra and Xa e =∑Xa . Now we prove the other inclusion. Take x ∈ lS rS (a), and n write 1 = i=1 ai ebi for some ai , bi in R. Then for any y ∈ rR (a), we get aeyai e = ayai e = 0 for∑each i. This implies that xeyai e = 0 for each i, which n gives xy = xey = xey i=1 ai ebi = 0 since x ∈ S. So x ∈ lR rR (a), and hence lS rS (a) ⊆ lR rR (a). Take x = s+t, where s ∈ Ra and t ∈ Xa . Hence, x = exe = ese + ete ∈ eRae + eXa e. This shows that lS rS (a) = eRae ⊕ eXa e = Sa ⊕ eXa , where eXa is a left ideal of S. Therefore, S is right AP S-injective. □ Remark 2.17. The condition that ReR = R in Proposition 2.16 is needed. For example, let R be the algebra of matrices, over a field F , of the form   a x 0 0 0 0  0 b 0 0 0 0     0 0 c y 0 0   . R=   0 0 0 a 0 0   0 0 0 0 b z  0 0 0 0 0 c By [5, Example 9], R is a QF -ring, and hence it is right AP S-injective. Let e = e11 + e22 + e44 + e55 be a sum of canonical matrix units. Then e is an idempotent of R such that ReR ̸= R and eRe ∼ = S = ( F0 F F ) . We claim that S is 0 F not right AP S-injective. In fact, J(S) = ( 0 0 ) . Then for any d = ( 00 d0 ) ∈ J(S), 0F lS rS (d) = ( 00 F F ) and Sd = ( 0 0 ) . So it does not exist a left ideal Xd of S such that lS rS (d) = Sd ⊕ Xd . Corollary 2.18. If the matrix ring Mn (R) over a ring R is right AP S-injective (n ≥ 1), then so is R. Proof. If S = Mn (R) is right AP S-injective, so is R ∼ = e11 Se11 by Proposition 2.16 because Se11 S = S (here eij is the matrix unit). □


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We do not know if the converse of Corollary 2.18 holds. However, we have the following result motivated by [11, Theorem 3.8]. Theorem 2.19. Let R be a ring and n ≥ 1. Then the following are equivalent: (1) Mn (R) is right AP S-injective. (2) HomR (Rn , R) is a direct summand of HomR (I, R) as left R-modules for any n-generated R-submodule I of J n . Proof. (1)⇒(2). Let S = Mn (R) and let I = a1 R + · · · + an R ∈ J n . Write (a1 , . . . , an ) = A, then A ∈ J(S). By hypothesis, we have lS rS (A) = SA ⊕ XA for some left ideal XA of S. Let     f (a1 ) · · · f (an )           0 · · · 0   Γ= f ∈ HomR (I, R) :  .  ∈ XA .. ..     . .       0 ··· 0 It is easy to verify that Γ is a left R-submodule of HomR (I, R). We claim that HomR (I, R) = HomR (Rn , R) ⊕ Γ as left R-modules. In fact, for any g ∈ HomR (I, R), write   g(a1 ) · · · g(an )  0 ··· 0    B= . . ..  ..  . ···

0

0

Then B ∈ lS rS (A), and∑hence B = (c ∑ijn)A + (dij ), where (cij ) ∈ S and (dij ) ∈ n XA . Let h : Rn → R, i=1 ei ri 7→∑ i=1 c1i ri , where ∑n ei is the standard basis n of Rn over R, and let k : I → R, i=1 ai ri 7→ i=1 d1i ri . Then g = h + k. Note that     d11 · · · d1n 1 ··· 0  0 ···   0     0 ··· 0   .. ..  =  .. ..  (dij ) ∈ XA .  . .   . .  0

···

0

0

···

0

So k ∈ Γ. Therefore, we have HomR (I, R) = HomR (Rn , R) + Γ. Suppose l ∈ HomR (Rn , R) ∩ Γ. Then there exists (c1 , . . . , cn ) ∈ Rn such that (l(a1 ), . . ., l(an )) = (c1 , . . . , cn )A. Thus,     l(a1 ) · · · l(an ) c1 · · · cn  0   ··· 0     0 ··· 0   .. ..  =  .. ..  A ∈ SA ∩ XA = 0.  . .   . .  0

···

0

0

···

Therefore, HomR (I, R) = HomR (Rn , R) ⊕ Γ.

0

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(2)⇒(1). Suppose A = (aij ) ∈ J(S). Let I = a1 R + · · · + an R, where ai is i-th column of A. Then I ∈ J n . By hypothesis, we have HomR (I, R) = HomR (Rn , R) ⊕ Γ for some left R-submodule Γ of HomR (I, R). Let    f1 (a1 ) · · · f1 (an )        f2 (a1 ) · · · f2 (an )     XA =   : fi ∈ Γ, i = 1, 2, . . . , n . .. ..     . .       fn (a1 ) · · · fn (an ) Then XA is a left ideal of S. Now we show that lS rS (A) = SA ⊕ XA as left S-modules. It is easy to check that XA ⊆ lS rS (A). If B = (bij ) ∈ lS rS (A), then rS (A) ⊆ rS (B). So f : AS → BS, A(sij ) 7→ B(sij ), (sij ) ∈ S is which induces an R-homomorphism fi : ∑na well-defined ∑S-homomorphism, n a r → 7 b r from I to R for each 1 ≤ i ≤ n. Write fi = j=1 j j j=1 ij j gi + hi , where gi ∈ HomR (Rn , R) and hi ∈ Γ. Then, for each i, there exists (ci1 , . . . , cin ) ∈ Rn such that (gi (a1 ), . . . , gi (an )) = (ci1 , . . . , cin )A. So,   h1 (a1 ) · · · h1 (an )  h2 (a1 ) · · · h2 (an )    B = (bij ) = (cij )A +   ∈ SA + XA , .. ..   . . hn (a1 )

···

hn (an )

showing lS rS (A) = SA + XA . Let C ∈ SA ∩ XA . Then for some (dij ) ∈ S and some ki ∈ Γ(i = 1, 2, . . . , n),   k1 (a1 ) · · · k1 (an )  k2 (a1 ) · · · k2 (an )    C=  ∈ (dij )A. .. ..   . . kn (a1 )

···

kn (an )

Then, for each i, (ki (a1 ), . . . , ki (an )) = (di1 , . . . , din )A, which shows that ki ∈ HomR (Rn , R) ∩ Γ = 0. Thus, each ki = 0, and hence C = 0. Therefore, lS rS (A) = SA ⊕ XA . □ The following theorem is a generalization of [17, Theorem 2.1]. Theorem 2.20. Let R be a right Noetherian, left AP S-injective ring. Then (1) lR (J) ≤e R R. (2) J is nilpotent. (3) lR (J) ≤e RR . Proof. (1) For any 0 ̸= x ∈ R, it is enough to show that lR (J) ∩ Rx ̸= 0. Since R has ACC on right annihilators, choose y ∈ R such that yx ̸= 0 and rR (yx) is maximal in {rR (ax)|a ∈ R, ax ̸= 0}. Now we prove that yxJ = 0. Otherwise, there exists a t ∈ J such that yxt ̸= 0. Note that yxt ∈ J and R is left AP S-injective, then rR lR (yxt) = yxtR ⊕ X for some right ideal X of R. We proceed with the following two cases.


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Case 1. rR lR (yx) = rR lR (yxt). Then yx ∈ rR lR (yxt) = yxtR ⊕ X. Write yx = yxtr+z, where yxtr ∈ yxtR−X and z ∈ X −yxtR. So z = yx(1−tr), and hence yx = z(1 − tr)−1 since 1 − tr is invertible, contradicting with yxtr ∈ / X. Case 2. rR lR (yx) ̸= rR lR (yxt). Then lR (yx) ̸= lR (yxt). It follows that there exists u ∈ lR (yxt) but u ∈ / lR (yx). Thus uyxt = 0 and uyx ̸= 0. This gives that t ∈ rR (uyx) and t ∈ / rR (yx). So rR (yx) ⊂ rR (uyx), contradicting the maximality of rR (yx). Then yxJ = 0, and so 0 ̸= yx ∈ lR (J) ∩ Rx. Therefore, lR (J) ≤e R R. (2) There exists k ≥ 1 such that lR (J k ) = lR (J k+1 ) = · · · . If J is not nilpotent, choose rR (x) to be maximal in {rR (y)|yJ k ̸= 0}. Then xJ 2k ̸= 0 because lR (J 2k ) = lR (J k ), so there exists b ∈ J k with xbJ k ̸= 0. Since lR (J) ≤ lR (J k ), we have lR (J k ) ≤e R R by (1). Thus, Rxb ∩ lR (J k ) ̸= 0, say 0 ̸= cxb ∈ lR (J k ). Hence, rR (x) ⊂ rR (cx) because xbJ k ̸= 0, contradicting the maximality of rR (x). (3) If 0 ̸= d ∈ R, we must show that dR ∩ lR (J) ̸= 0. It is clear if dJ = 0. Otherwise, since J is nilpotent by (2), there exists m ≥ 1 such that dJ m ̸= 0 but dJ m+1 = 0. Then 0 ̸= dJ m ⊆ dR ∩ lR (J), as desired. □ In [17], a module M is said to satisfy the generalized C2-condition (GC2) if, for any N ⊆ M and N ∼ = M , N is a summand of M . Corollary 2.21. If R is a right Noetherian, left AP S-injective ring such that RR satisfies (GC2), then it is right Artinian. Proof. Note that R is right finitely dimensional. By [17, Lemma 1.1], R is semilocal. By Theorem 2.18, J(R) is nilpotent. Thus, R is semiprimary, and hence it is right Artinian by the Hopkins-Levitzki theorem. □ The condition that RR satisfies (GC2) can not be omitted. For example, the ring R = Z is a Noetherian and AP S-injective ring but not Artinian. From [10, Proposition 1.46], a left Kasch ring is right C2. Thus, we have the following corollary. Corollary 2.22. If R is a right Noetherian, left Kasch and left AP S-injective ring, then it is right Artinian. 3. Trivial extensions Let R be a ring and M a bimodule over R. The trivial extension of R and M is R ∝ M = {(a, x) | a ∈ R, x ∈ M } with addition defined componentwise and multiplication defined by (a, x)(b, y) = (ab, ay + xb). For convenience, we write I ∝ X = {(a, x) | a ∈ I, x ∈ X}, where I is a subset of R and X is a subset of M . It is easy to check that J(R ∝ M ) = J(R) ∝ M .

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Proposition 3.1. Let R be a ring and Xa a left ideal of R for any a ∈ R, S = R ∝ R. Then the following are equivalent: (1) lR rR (a) = Ra ⊕ Xa . (2) lS rS (0, a) = S(0, a) ⊕ X(0,a) , where X(0,a) = 0 ∝ Xa is a left ideal of S. (3) lS rS (a, 0) = S(a, 0) ⊕ X(a,0) , where X(a,0) = Xa ∝ 0 is a left ideal of S. (4) lS rS (a, a) = S(a, a) ⊕ X(a,a) , where X(a,a) = Xa ∝ Xa is a left ideal of S. Proof. (1)⇒(2). For any (b, c) ∈ lS rS (0, a), rS (0, a) ⊆ rS (b, c). Since (0, 1) ∈ rS (0, a), 0 = (b, c)(0, 1) = (0, b), showing b = 0. If x ∈ rR (a), then (x, 0) ∈ rS (0, a) ⊆ rS (b, c), showing that 0 = (0, c)(x, 0) = (0, cx). So x ∈ rR (c), and hence rR (a) ⊆ rR (c). Thus, c ∈ lR rR (c) ⊆ lR rR (a) = Ra ⊕ Xa . Write c = ra + y, where ra ∈ Ra − Xa and y ∈ Xa − Ra. Then (b, c) = (0, ra + y) = (r, 0)(0, a) + (0, y) ∈ S(0, a) + X(0,a) , where X(0,a) = 0 ∝ Xa is a left ideal of S. It is easy to prove that S(0, a) ∩ X(0,a) = 0, so lS rS (0, a) ⊆ S(0, a) ⊕ X(0,a) . Conversely, for any (m, n) ∈ S(0, a) ⊕ X(0,a) , where X(0,a) = 0 ∝ Xa is a left ideal of S. Then (m, n) = (r1 , r2 )(0, a) + (0, y) = (0, r1 a + y), where (r1 , r2 )(0, a) ∈ S(0, a) − X(0,a) and (0, y) ∈ X(0,a) − S(0, a). Note that r1 a ∈ Ra − Xa and y ∈ Xa − Ra, so m = 0, n = r1 a + y ∈ Ra ⊕ Xa = lR rR (a). Then rR (a) ⊆ rR (n). For any (k, l) ∈ rS (0, a), 0 = (0, a)(k, l) = (0, ak), showing k ∈ rR (a), and hence nk = 0. Then (m, n)(k, l) = (0, n)(k, l) = (0, nk) = 0, proving (m, n) ∈ lS rS (0, a). (2)⇒(1). For any b ∈ lR rR (a), rR (a) ⊆ rR (b). If (x, y) ∈ rS (0, a), then ax = 0. So x ∈ rR (a) ⊆ rR (b), showing (0, b)(x, y) = 0. Thus, (x, y) ∈ rS (0, b). So rS (0, a) ⊆ rS (0, b). This gives that (0, b) ∈ lS rS (0, b) ⊆ lS rS (0, b) = S(0, a) ⊕ X(0,a) . Write (0, b) = (r1 , r2 )(0, a) + (0, y) = (0, r1 a + y), where (r1 , r2 )(0, a) ∈ S(0, a) − X(0,a) and (0, y) ∈ X(0,a) − S(0, a). Note that r1 a ∈ Ra−Xa and y ∈ Xa −Ra, so b = r1 a+y ∈ Ra⊕Xa , proving lR rR (a) ⊆ Ra⊕Xa . Now we show the other inclusion. For any c ∈ Ra ⊕ Xa , write c = ra + z, where ra ∈ Ra − Xa and z ∈ Xa − Ra. Then (0, c) = (0, ra) + (0, z) = (r, 0)(0, a) + (0, z) ∈ S(0, a) ⊕ X(0,a) = lS rS (0, a). So rS (0, a) ⊆ rS (0, c). If x ∈ rR (a), then (x, 0) ∈ rS (0, a), showing 0 = (0, c)(x, 0) = (0, cx), and hence x ∈ rR (c). Thus, rR (a) ⊆ rR (c). This implies that c ∈ lR rR (c) ⊆ lR rR (a). The proofs of (1)⇔(3) and (1)⇔(4) are similar to that of (1)⇔(2). □ Corollary 3.2. Let R be a ring and a ∈ R, S = R ∝ R. Then the following are equivalent: (1) (2) (3) (4)

lR rR (a) = Ra. lS rS (0, a) = S(0, a). lS rS (a, 0) = S(a, 0). lS rS (a, a) = S(a, a).


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Corollary 3.3. Let R be a ring. If R ∝ R is right AP S-injective, then R is right AP -injective. Proof. Let S = R ∝ R. For any 0 ̸= a ∈ R, (0, a) ∈ J(S). So there exists a left ideal X(0,a) of S such that lS rS (0, a) = S(0, a) ⊕ X(0,a) . By the proof of (1)⇒(2) in Proposition 3.1, if (b, c) ∈ lS rS (0, a) and (m, n) ∈ S(0, a), then b = 0 and m = 0. So X(0,a) = 0 ∝ Xa , where Xa is a left ideal of R. By Proposition 3.1 again, we have lR rR (a) = Ra ⊕ Xa , proving that R is right AP -injective. □ Remark 3.4. We claim that R being right AP S-injective can not imply R ∝ R being right AP S-injective. For example, let R = Z be the ring of integers. Suppose that S = Z ∝ Z is AP S-injective, then Z is AP -injective by Corollary 3.3, a contradiction. For f, g ∈ HomR (R, R), define α = (f, g) such that α(a, b) = (f, g)(a, b) = (f (a), f (b) + g(a)), where (a, b) ∈ S = R ∝ R. It is easy to check that α ∈ HomS (S, S). Conversely, for any α ∈ HomS (S, S), let α(1, 0) = (p, q). Define f (1) = p, g(1) = q, then f, g ∈ HomR (R, R) and α = (f, g). In the following theorem, we shall discuss when R ∝ R is right AP S-injective. Theorem 3.5. Let R be a ring. If, for any a ∈ J(R), b ∈ R, HomR (aR + brR (a), R) = HomR (R, R) ⊕ X as left R-modules for some submodules X of HomR (aR + brR (a), R), then R ∝ R is right AP S-injective. Proof. Let S = R ∝ R and any A ∈ J(S). By Lemma 2.2, it is enough to show that HomS (AS, S) = HomS (S, S) ⊕ Y for some left S-submodules Y of HomS (AS, S). Write A = (a, b), then a ∈ J(R), b ∈ R. For any f ∈ HomS (AS, S), say f (A) = (p, q), p, q ∈ R. Define g : aR + brR (a) → R, ar1 + br2 7→ pr1 + qr2 . If ar1 + br2 = 0, then (a, b)(r2 , r1 ) = (ar2 , ar1 + br2 ) = 0 since ar2 = 0, and hence 0 = f ((a, b)(r2 , r1 )) = (p, q)(r2 , r1 ) = (pr2 , pr1 + qr2 ), which implies pr1 + qr2 = 0. So g ∈ HomR (aR + brR (a), R). By hypothesis, g = h ⊕ k, where h ∈ HomR (R, R) and k ∈ X. In particular, p = g(a) = h(a) + k(a) = h(1)a + k(a). If a = 0, then rR (a) = R. So R is right AP -injective. Define l : aR → R, ar 7→ qr − h(1)br − k(br), r ∈ R. If ar = 0, then r ∈ rR (a), so h(1)br = h(br) = g(br) − k(br) = qr − k(br), and hence qr − h(1)br − k(br) = 0. Thus, l ∈ HomR (aR, R). Then l = h′ ⊕ k ′ , where h′ ∈ HomR (R, R) and k ′ ∈ K ⊆ HomR (aR, R). We have q − h(1)b − k(b) = l(a) = h′ (a) + k ′ (a), so q = h(1)b + k(b)+h′ (1)a+k ′ (a). Then f (A) = (p, q) = (h(1)a+k(a), h(1)b+k(b)+h′ (1)a+ k ′ (a)) = (h(1), h′ (1))(a, b) + (k(a), k(b) + k ′ (a)) = (h, h′ )(a, b) + (k, k ′ )(a, b) = (h, h′ )A+(k, k ′ )A. Note that (h, h′ ) ∈ HomS (S, S) and (k, k ′ ) ∈ Y = {(i, j)|i ∈ X, j ∈ K} ⊆ HomS (AS, S), then HomS (AS, S) ⊆ HomS (S, S) + Y . Now we show that HomS (S, S) ∩ Y = 0. For any y ∈ HomS (S, S) ∩ Y , write y = (m, n), then m ∈ HomR (R, R) ∩ X = 0 and n ∈ HomR (R, R) ∩ K = 0, proving y = 0. Therefore, HomS (AS, S) ⊆ HomS (S, S) ⊕ Y . But the other inclusion is clear, as desired. □

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Following the preceding theorem, we immediately deduce the following corollaries. Corollary 3.6. Let R be a ring. If, for any a, b ∈ R, HomR (aR+brR (a), R) = HomR (R, R) ⊕ X as left R-modules for some submodules X of HomR (aR + brR (a), R), then R ∝ R is right AP -injective. Corollary 3.7. Let R be a ring. If, for any a ∈ J(R), b ∈ R, any Rhomomorphism aR + b · r(a) → R can be extended to R, then R ∝ R is right P S-injective. Corollary 3.8. If R is semiprimitive and right AP -injective, then R ∝ R is right AP S-injective. Remark 3.9. As Remark 3.4, the condition that R is right AP -injective in Corollary 3.8 can not be omitted.

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Yueming Xiang Department of Mathematics and Applied Mathematics Huaihua University Huaihua, 418000, P. R. China E-mail address: [email protected]

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